\(\dfrac{1}{\sqrt{x}-2}+\dfrac{2}{\sqrt{X}+2}-\dfrac{2\sqrt{X}}{x-4}\)<...">
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14 tháng 12 2017

\(P=\dfrac{1}{\sqrt{x}-2}+\dfrac{2}{\sqrt{x}+2}-\dfrac{2\sqrt{x}}{x-4}\)

a) \(ĐKXĐ:x\ge0;x\ne4\)

b) \(P=\dfrac{1}{\sqrt{x}-2}+\dfrac{2}{\sqrt{x}+2}-\dfrac{2\sqrt{x}}{x-4}\)

\(=\dfrac{1}{\sqrt{x}-2}+\dfrac{2}{\sqrt{x}+2}-\dfrac{2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{\sqrt{x}+2+2\left(\sqrt{x}-2\right)-2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{\sqrt{x}+2+2\sqrt{x}-4-2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{1}{\sqrt{x}+2}\)

Vậy: \(P=\dfrac{1}{\sqrt{x}+2}\)

_Chúc bạn học tốt_

13 tháng 12 2020

a) ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)

Ta có: \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right):\dfrac{2\sqrt{x}}{x-4}\)

\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\cdot\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right):\dfrac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\left(\dfrac{x+2\sqrt{x}+x-2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\dfrac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{2x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{2\sqrt{x}}\)

\(=\sqrt{x}\)

b) Để P>4 thì \(\sqrt{x}>4\)

hay x>16

Kết hợp ĐKXĐ, ta được: x>16

Vậy: Khi x>16 thì P>4

13 tháng 12 2020

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27 tháng 12 2018

Câu 1 chưa rõ đề !

Câu 2 :

a ) ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)

b ) \(P=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{2+5\sqrt{x}}{4-x}\)

\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{2+5\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)

c ) \(P=2\Leftrightarrow\dfrac{3\sqrt{x}}{\sqrt{x}+2}=2\Leftrightarrow\sqrt{x}=4\Leftrightarrow x=16\)

27 tháng 12 2018

2, a,ĐKXĐ:\(\left\{{}\begin{matrix}\sqrt{x}\ge0\\\left\{{}\begin{matrix}\sqrt{x}-2\ne0\\\sqrt{x}+2\ne0\\4-x\ne0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
b,\(P=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{2+5\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)\(P=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)+2\sqrt{x}\left(\sqrt{x}-2\right)-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(P=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(P=\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(P=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
c, P=2\(\Leftrightarrow\dfrac{3\sqrt{x}}{\sqrt{x}+2}=2\)
\(\Leftrightarrow3\sqrt{x}=2\left(\sqrt{x}+2\right)\Leftrightarrow\sqrt{x}=4\)
\(\Leftrightarrow x=16\)
Vậy x=16 thì P có giá trị =2

12 tháng 8 2021

a) ĐKXĐ: \(x\ge0;x\ne1\)

b) \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\right):\dfrac{2}{\sqrt{x}+1}\left(x\ge0;x\ne1\right)\\ P=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{2}\\ P=\dfrac{x-\sqrt{x}-x-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{2}\\ P=\dfrac{-2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{2}\\ P=\dfrac{-\sqrt{x}}{\sqrt{x}-1}\)

12 tháng 8 2021

Giúp mình với

14 tháng 7 2018

a) ĐKXĐ: x ≥ 0, x # 9

(quá trình thì b tự làm nha)

b) Ta có:

\(A=\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{7\sqrt{x}+4}{x-\sqrt{x}-6}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\)

= \(\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+7\sqrt{x}+4-\left(\sqrt{x}+2\right)^2}{\left(\sqrt{x}+2\right)\left(\sqrt{x-3}\right)}\)

= \(\dfrac{2x-3\sqrt{x}+7\sqrt{x}+4-x-4\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)

= \(\dfrac{x-3\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)

= \(\dfrac{\sqrt{x}}{\sqrt{x}+2}\)

Vậy A = \(\dfrac{\sqrt{x}}{\sqrt{x}+2}\) với x ≥ 0, x # 9

c) ĐKXĐ: x ≥ 0, x # 9, x ∈ Z

Ta có: \(A=\dfrac{\sqrt{x}}{\sqrt{x}+2}=\dfrac{\sqrt{x}+2-2}{\sqrt{x}+2}=1-\dfrac{2}{\sqrt{x}+2}\)

Để A có gtn ⇔ \(1-\dfrac{2}{\sqrt{x}+2}\) nguyên ⇔ \(\dfrac{2}{\sqrt{x}+2}\) nguyên

\(\sqrt{x}+2\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

\(\sqrt{x}+2\in\left\{1;2\right\}\) (Vì \(\sqrt{x}+2>0\))

Nếu \(\sqrt{x}+2=1\)thì \(\sqrt{x}=-1\) (Vô nghiệm)

Nếu \(\sqrt{x}+2=2\) thì \(\sqrt{x}=0\Leftrightarrow x=0\)(TMĐK)

Vậy để A có gtn thì x = 0

1 tháng 8 2018

a/ đkxđ: x > 0; x≠1

b/ \(A=\left(\dfrac{\sqrt{x}}{2}-\dfrac{1}{2\sqrt{x}}\right):\left(\dfrac{x-\sqrt{x}}{\sqrt{x}+1}-\dfrac{x+\sqrt{x}}{\sqrt{x}-1}\right)\)

\(=\dfrac{x-1}{2\sqrt{x}}\cdot\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)^2-\sqrt{x}\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)

\(=\dfrac{x-1}{2\sqrt{x}}\cdot\dfrac{x\sqrt{x}-2x+\sqrt{x}-x\sqrt{x}-2x-\sqrt{x}}{x-1}\)

\(=\dfrac{-4x}{2\sqrt{x}}=-2\sqrt{x}\)

1 tháng 8 2018

c/ A > -6

\(\Leftrightarrow-2\sqrt{x}>-6\Leftrightarrow\sqrt{x}< 3\Leftrightarrow x< 9\)

kết hợp với đkxđ => 0 < x < 9

25 tháng 12 2017

a.

\(P=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{2+5\sqrt{x}}{4-x}\\ =\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{x-4}+\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{x-4}-\dfrac{2+5\sqrt{x}}{x-4}\\ =\dfrac{x+\sqrt{x}-2+x-4\sqrt{x}}{x-4}\\ =\dfrac{2x-2-3\sqrt{x}}{x-4}\)

Bài 1: 

a: \(B=\dfrac{\sqrt{x}+x+\sqrt{x}-x}{1-x}\cdot\dfrac{x-1}{3-\sqrt{x}}\)

\(=\dfrac{2\sqrt{x}}{\sqrt{x}-3}\)

b: Để B=-1 thì \(2\sqrt{x}=-\sqrt{x}+3\)

=>3 căn x=3

=>căn x=1

hay x=1(loại)

18 tháng 3 2021

a, Với \(x>0;x\ne4;x\ne9\)

\(A=\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{4-x}\right):\left(\frac{\sqrt{x}-1}{x-2\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\)

\(=\left(\frac{4\sqrt{x}\left(2-\sqrt{x}\right)+8x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)

\(=\left(\frac{8\sqrt{x}-4x+8x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)

\(=\frac{8\sqrt{x}+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}:\frac{-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\frac{4\sqrt{x}\left(2+\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\frac{3-\sqrt{x}}{\sqrt{x}\left(2-\sqrt{x}\right)}=\frac{4\sqrt{x}}{2-\sqrt{x}}.\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{3-\sqrt{x}}=\frac{4x}{3-\sqrt{x}}\)

b, Ta có : A = -2 hay 

\(\frac{4x}{3-\sqrt{x}}=-2\Rightarrow4x=-6+2\sqrt{x}\)

\(\Leftrightarrow4x+6-2\sqrt{x}=0\Leftrightarrow2\left(2x+3-\sqrt{x}\right)=0\)

\(\Leftrightarrow2x+3-\sqrt{x}=0\Leftrightarrow\sqrt{x}=2x+3\)

bình phương 2 vế ta có : 

\(x=\left(2x+3\right)^2=4x^2+12x+9\)

\(\Leftrightarrow-4x^2-11x-9=0\)giải delta ta thu được : \(x=-\frac{11\pm\sqrt{23}i}{8}\)

\(a,A=\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{4-x}\right):\left(\frac{\sqrt{x}-1}{x-2\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\)              

\(=\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\frac{2}{\sqrt{x}}\right)\)

\(=\frac{4\sqrt{x}.\left(2-\sqrt{x}\right)+8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\frac{\sqrt{x}-1-2.\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\frac{8\sqrt{x}-4x+8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-1-2\sqrt{x}+4}\)

\(=\frac{\left(4x+8\sqrt{x}\right)\left(\sqrt{x}\right)\left(\sqrt{x}-2\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)\left(-\sqrt{x}+3\right)}\)

\(=\frac{-4\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}\right)\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)\left(-\sqrt{x}+3\right)}\)

\(=\frac{4x}{\sqrt{x}-3}\)

30 tháng 1 2019

a) ĐKXĐ: \(x\ge0;x\ne16\)

b) \(\left(\dfrac{\sqrt{x}}{\sqrt{x}+4}+\dfrac{4}{\sqrt{x}-4}\right):\dfrac{x+16}{\sqrt{x}+2}\)

\(\Leftrightarrow\dfrac{\sqrt{x}\left(\sqrt{x}-4\right)+4\left(\sqrt{x}+4\right)}{x-16}:\dfrac{x+16}{\sqrt{x}+2}\)

\(\Leftrightarrow\dfrac{x-4\sqrt{x}+4\sqrt{x}+16}{x-16}.\dfrac{\sqrt{x}+2}{x+16}\)

\(\Leftrightarrow\dfrac{\sqrt{x}+2}{x-16}\)

30 tháng 1 2019

a) ĐKXĐ: \(x\ge0,x\ne16\)

b) \(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}+4}+\dfrac{4}{\sqrt{x}-4}\right):\dfrac{x+16}{\sqrt{x}+2}\)

\(B=\dfrac{\left(\sqrt{x}-4\right)\sqrt{x}+4\left(\sqrt{x}+4\right)}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}.\dfrac{\sqrt{x}+2}{x+16}\\ B=\dfrac{x-4\sqrt{x}+4\sqrt{x}+16}{x-16}.\dfrac{\sqrt{x}+2}{x+16}\\ B=\dfrac{1}{x-16}.\left(\sqrt{x}+2\right)\\ B=\dfrac{\sqrt{x}+2}{x-16}\)